The argument:

While building a mount for a band saw motor, a friend of mine started to argue over the trig involved to get the most tension in the belt, given a fixed weight of the motor and lack of interest in adding bricks.

After over a day of arguing, I'm sick of arguing. So I though id let a forum of machinists/engineers/highly intelligent people do it for me.

We have drawn a set of figures, and writing out conclusions on what actually supplies the belt tension.

Each figure, has pivot points connecting each segment. The top segment is considered a belt/rope, and as such is always held in tension by the lower segment, a rigid bar.

The two segments are assumed to be weightless, with only the weight affixed to the end to simplify things.

The lower bar is allways the same length for the sake of simplicity and eliminating another variable.

We where both given this picture to rate from highest belt tension to lowest:

http://www.voxelsoft.com/spacesim/pulleys2.png

My conclusion:

Its a simple matter of leverage advantage.

The smaller the angle A, The less the top belt would strech per inch of the motor droped

Any time you trade off distance, you gain force.

The lever nature can be more easily seen after altering it slightly, as shown in my picture, with the new line being fixed in angle to the bottom arm, but the pivot point of the fulcrum being the lower left point. Basicly a 'bent' lever

http://www.voxelsoft.com/spacesim/pulleys3.png

As such: Tension on the belt highest with a small angle A (Higher leverage advantage) and angle B to be horizontal (mass furthest out)

Assuming angle B translated into a value of 1 when horizontal, and 0 when vertical

And angle A translated into a value of infinity at 0 (parallel), and 1 when at 90 degrees. I don't actually feel like doing the proper trig for this but im just looking for a

very rough answer here. I know there's various trig functions to compute the actual vector of forces here.

The tension on the top arm (or belt) is similar to Weight * A * B

As such, both angles are equally important in producing the highest belt tension. With 90 degrees of A, you only get a max weight of 1, With near 90 degrees of B, you get

nearly no weight to multiply by leverage advantage.

The figures with the most tension would be from highest to lowest:

5

6 tied with 7

1, 3, 2, 4 all very close together, But very far off from 5, 6 and 7.

As far as the 'primary source' of the tension, Yes B must not be near 90 degrees, but generally, A will preferably be small in a band saw motor mount, and multiply the force by

many times.

I showed him a physical model of the setup, Showed him the tension on a rope increases as angle A was reduced until he was unable to even lift a 10lb weight by pulling on the string. He still does not believe me.

http://www.voxelsoft.com/spacesim/pulleys4.png

Thats his explantion.

While building a mount for a band saw motor, a friend of mine started to argue over the trig involved to get the most tension in the belt, given a fixed weight of the motor and lack of interest in adding bricks.

After over a day of arguing, I'm sick of arguing. So I though id let a forum of machinists/engineers/highly intelligent people do it for me.

We have drawn a set of figures, and writing out conclusions on what actually supplies the belt tension.

Each figure, has pivot points connecting each segment. The top segment is considered a belt/rope, and as such is always held in tension by the lower segment, a rigid bar.

The two segments are assumed to be weightless, with only the weight affixed to the end to simplify things.

The lower bar is allways the same length for the sake of simplicity and eliminating another variable.

We where both given this picture to rate from highest belt tension to lowest:

http://www.voxelsoft.com/spacesim/pulleys2.png

My conclusion:

Its a simple matter of leverage advantage.

The smaller the angle A, The less the top belt would strech per inch of the motor droped

Any time you trade off distance, you gain force.

The lever nature can be more easily seen after altering it slightly, as shown in my picture, with the new line being fixed in angle to the bottom arm, but the pivot point of the fulcrum being the lower left point. Basicly a 'bent' lever

http://www.voxelsoft.com/spacesim/pulleys3.png

As such: Tension on the belt highest with a small angle A (Higher leverage advantage) and angle B to be horizontal (mass furthest out)

Assuming angle B translated into a value of 1 when horizontal, and 0 when vertical

And angle A translated into a value of infinity at 0 (parallel), and 1 when at 90 degrees. I don't actually feel like doing the proper trig for this but im just looking for a

very rough answer here. I know there's various trig functions to compute the actual vector of forces here.

The tension on the top arm (or belt) is similar to Weight * A * B

As such, both angles are equally important in producing the highest belt tension. With 90 degrees of A, you only get a max weight of 1, With near 90 degrees of B, you get

nearly no weight to multiply by leverage advantage.

The figures with the most tension would be from highest to lowest:

5

6 tied with 7

1, 3, 2, 4 all very close together, But very far off from 5, 6 and 7.

As far as the 'primary source' of the tension, Yes B must not be near 90 degrees, but generally, A will preferably be small in a band saw motor mount, and multiply the force by

many times.

I showed him a physical model of the setup, Showed him the tension on a rope increases as angle A was reduced until he was unable to even lift a 10lb weight by pulling on the string. He still does not believe me.

http://www.voxelsoft.com/spacesim/pulleys4.png

Thats his explantion.

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